= Inertia tensor
{title2=$I_{ij}$}
{wiki}
The <inertia tensor> packages the <moments of inertia> about every axis through a specified point $O$ into one <symmetric matrix>. An axis still needs a direction: if $\mathbf n$ is a <unit vector>, its scalar <moment of inertia> is $I_{O,\mathbf n}=\mathbf n^T\mathbf I_O\mathbf n$. The same <mass> layout can therefore be easy to spin about one direction and harder about another. The <inertia tensor> also determines <rotational kinetic energy> and relates <angular velocity> to <angular momentum> in the appropriate <rigid body> model.
Take <position> vectors $\mathbf r=(x,y,z)$ relative to $O$, expressed in an <orthonormal basis>. For <point masses> $m_a$, or a continuous distribution with <mass density> $\rho$ and $dm=\rho\,dV$,
$$
(\mathbf I_O)_{ij}=\sum_a m_a\bigl(|\mathbf r_a|^2\delta_{ij}-r_{ai}r_{aj}\bigr),
\qquad
(\mathbf I_O)_{ij}=\int\bigl(|\mathbf r|^2\delta_{ij}-r_ir_j\bigr)\,dm.
$$
Here $\delta_{ij}$ is the <Kronecker delta>. Equivalently, using the <identity matrix> $\mathbf1$ and an <outer product>,
$$
\mathbf I_O=\int\bigl(|\mathbf r|^2\mathbf1-\mathbf r\mathbf r^T\bigr)\,dm
=\int\begin{pmatrix}
y^2+z^2&-xy&-xz\\
-xy&x^2+z^2&-yz\\
-xz&-yz&x^2+y^2
\end{pmatrix}\,dm.
$$
Every entry has <SI> units $\mathrm{kg\,m^2}$. The diagonal entries are axial <moments of inertia>; the off-diagonal entries carry a minus sign in this convention. Some texts instead define the integrals without the prefixed minus sign $\int xy\,dm$, $\int xz\,dm$, $\int yz\,dm$ as their named off-diagonal quantities and put the minus signs into the displayed <matrix>. Check that convention before combining formulas. <Newtonian Dynamics by Richard Fitzpatrick/Moment of Inertia Tensor>{ref} <16.07 Dynamics/Lecture L26 - 3D Rigid Body Dynamics: The Inertia Tensor>{ref}
The <symmetric matrix> is <positive semidefinite>, because its <quadratic form> is
$$
\mathbf u^T\mathbf I_O\mathbf u
=\int\bigl(|\mathbf u|^2|\mathbf r|^2-(\mathbf u\cdot\mathbf r)^2\bigr)\,dm
=\int|\mathbf u\times\mathbf r|^2\,dm\geq0.
$$
The <dot product> and <cross product> identify this as a sum of squared distances when $\mathbf u$ is a <unit vector>. A zero <moment of inertia> occurs precisely when all the <mass> lies on that axis, up to sets of zero <mass>. A genuine three-dimensional body with positive <mass density> in a volume has strictly positive axial <moments of inertia>. Do not confuse this <inertia tensor> with the <mass second-moment tensor> $\mathbf S=\int\mathbf r\mathbf r^T\,dm$: they obey $\mathbf I_O=\operatorname{tr}(\mathbf S)\mathbf1-\mathbf S$, where $\operatorname{tr}$ denotes the <matrix trace>.
If $O$ is fixed in an <inertial frame> and each <velocity> is $\mathbf v=\boldsymbol\omega\times\mathbf r$, summing the individual <kinetic energies> and <angular momenta> gives
$$
\mathbf L_O=\mathbf I_O\boldsymbol\omega,
\qquad
K=\frac12\boldsymbol\omega^T\mathbf I_O\boldsymbol\omega
=\frac12\boldsymbol\omega\cdot\mathbf L_O.
$$
These formulas also describe <angular momentum> and <rotational kinetic energy> relative to the <centre of mass> $G$ of a translating <rigid body>. In that case the total <kinetic energy> in the <inertial frame> is $K_{\mathrm{total}}=\tfrac12M|\mathbf V_G|^2+\tfrac12\boldsymbol\omega^T\mathbf I_G\boldsymbol\omega$. For a general moving point $O$, the body's actual <velocities> include $\mathbf V_O$, so $\mathbf I_O\boldsymbol\omega$ alone need not be its <angular momentum> about $O$. <Newtonian Dynamics by Richard Fitzpatrick/Rotational Kinetic Energy>{ref} <16.07 Dynamics/Lecture L26 - 3D Rigid Body Dynamics: The Inertia Tensor>{ref}
Changing the <orthonormal basis> changes components, rather than the physical distribution: if the columns of an <orthogonal matrix> $Q$ are the new basis vectors expressed in the old basis, then $\mathbf I'_O=Q^T\mathbf I_OQ$. Changing the reference point is a different operation, governed by the <tensor parallel-axis theorem>. Both <principal axes> and <principal moments of inertia> must therefore be quoted with their reference point.
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